Question: What is the area of the region between the graphs of $f(x)=\sqrt{x+1}$ and $g(x)=2x-4$ from $x=0$ to $x=3$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-3$ (Choice B) B $\dfrac{14}{3}$ (Choice C) C $\dfrac{5}{3}$ (Choice D) D $\dfrac{23}{3}$
Explanation: Visualizing the area We sketch the graphs of $f$ and $g$ first. ${1}$ ${2}$ ${3}$ ${2}$ ${\llap{-}2}$ ${\llap{-}4}$ $f$ $g$ $y$ $x$ From the graph, it appears that $f(x)\ge g(x)$ between $x=0$ and $x=3$. From this we are looking to evaluate: $ \int_{0}^{3}\left( f(x)-g(x) \right)\,dx$ Evaluating the definite integral $\begin{aligned} &\phantom{=} \int_{0}^{3} \left( \sqrt{x+1}- (2x-4) \right) \,dx \\\\ &= \dfrac{2(x+1)^{3/2}}{3} -x^2 +4x ~\Bigg|_{0}^{3} \\\\ &= \left( \dfrac{16}{3} - 9 + 12 \right) -\left( \dfrac{2}{3} - 0 + 0 \right)\\\\ &= \dfrac{23}{3} \end{aligned}$ Answer The area is $\dfrac{23}{3}$ square units.